Ask question

Ask question

All categories

3 years ago

10

The Sun's angular diameter is measured to be 1920''.The distance D of the Sun from the Earth is 1.496×10^11 m.What is the diamet

er of the Sun?

1 answer:

insens350 [35]3 years ago

8 0

The Sun's angular diameter is measured to be

The distance of the sun from the earth is 1.496 × m.

You might be interested in

A boy yells in a narrow valley. The sound reflects off of a boulder that is 80 m away. How long will it take to hear the echo if

soldier1979 [14.2K]

Answer:4 lubeconds (speed=distance over time)

4 0

3 years ago

Read 2 more answers

A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5

Phoenix [80]

Answer:

The propulsion force at the moment of departure, is 49 N

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N

Therefore, the propulsion force at the moment of departure, is 49 N

7 0

4 years ago

Do you get more flexible by stretching

Alex Ar [27]

Answer:

Yes

Explanation:

By stretching, parts pf your body loosen and become more flexible.

3 0

3 years ago

Read 2 more answers

1. What does the SLOPE of a displacement vs time graph mean in terms of an object smo

Zarrin [17]

C.The acceleration of the object

6 0

3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago